The solution of $ x ^ 3 = q $ is to calculate the recurrence formula $x_ {n + 1} = x_n-\ frac {x_ {n} ^ 3 --q} {3x_ {n} ^ 2}$ It can be calculated approximately with.

Put a positive number $\ frac {q} {2}$ in $ x_1 $

$x_2 = x_1-\ frac {x_ {1} ^ 3 --q} {3x_ {1} ^ 2}$, $x_3 = x_2-\ frac {x_ {2} ^ 3 --q} {3x_ {2} ^ Calculate as 2}$,….

While doing this calculation

When the value of $ | x ^ 3 --q | $ becomes small enough, stop the calculation and use the last calculated $ x_n $ as an approximate solution of $ x ^ 3 = q $.

Follow this method to create a program that outputs an approximation of the cube root of $ q $ for the input positive integer $ q $. However, use $ | x ^ 3-q | <0.00001 q $ to determine that it is "small enough".

input

Multiple datasets are given. For each dataset, $ q $ ($1 \ leq q <2 ^ {31}$) (integer) is given on one line. The end of the input is -1.

The number of datasets does not exceed 50.

output

Print $ x $ (real number) on one line for each dataset. The output result may contain an error of 0.00001 or less.

Example

Input

15
15
-1

Output

2.466212
2.466212

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